Heat of solution of BaCl2.2H2O = 200 kJ mol–1Heat of hydration of BaCl2 = – 150 kJ mol–1 Hence heat of solution of BaCl2 is350 kJ mol–1 50 kJ mol–1 –350 kJ mol–1 250 kJ mol–1

What is the predicted change in the boiling point of water when 1.50 g of barium chloride (BaCl2) is dissolved in 1.50 kg of water?Kb of water = 0.51°C/molmolar mass BaCl2 = 208.23 g/moli value of BaCl2 = 3A.0.51°CB.-0.00082°CC.0.0073°CD.-0.0024°C

Some chlorides are hydrated.When hydrated barium chloride crystals, BaC*l_{j}*x*H_{j}*O are heated they give off water.BaC*l_{2}*x*H_{2}*O(s) -> BaC*l_{2}(s) + x*H_{2}*O(g)In an experiment, 4.88g of BaC*l_{2}*x*H_{2}*O is heated until all the water is given off. The mass of BaC*l_{2} remaining is 4.16g.[M. BaC*l_{D} 208 H_{2}*O, 18 ]Determine the value of x using the following steps. Calculate the number of moles of BaC*l_{2} remainingmolCalculate the mass of H_{2}*O given off.Calculate the number of moles of H_{2}*O given off. Determine the value of x.molx = ..[4]

ΔH for the following reaction is -431 kJ. How much heat is released when 200 g of Ba(s) are reacted with an excess of H2O(l)? The molar mass of Barium is 137.3 g/mol. Ba(s) + 2 H2O(l) –> Ba(OH)2(aq) + H2(g) Group of answer choices291 kJ191 kJ491 kJ391 kJ

Calculate the heat of the following reaction:2C2H2 + 5O2 → 4CO2 + 2H2OGiven the following heats of formation:C2H2                    ΔH°f= 226.7 kJ/molO2                       ΔH°f= 0.0 kJ/molCO2                    ΔH°f= -393.5 kJ/molH2O                    ΔH°f= -241.8 kJ/molGroup of answer choices0.0 kJ/mol-2511 kJ/mol-2284.3 kJ/mol408.6 kJ/mol

When 2.35g Mg(OH)2 is added to 250.0 mL of water, the temperature of the waterraises from 20.5oC to 36.0oC. Calculate the molar enthalpy of solution.